Find the constant term in the expansion of?

((-4x^6)+(-2/x))^35

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  • 7 years ago
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    Recall that by the Binomial Theorem:

    (a + b)^n = ∑ [C(n, k) * a^(n - k) * n^k] (from k=0 to n).

    So:

    [-4x^6 + (-2/x)]^35 = ∑ [C(35, k) * (-4x^6)^(35 - k) * (-2/x)^k] (from k=0 to 35).

    Then, using the laws of exponents to get x's exponent in terms of k:

    ∑ [C(35, k) * (-4x^6)^(35 - k) * (-2/x)^k] (from k=0 to 35)

    = ∑ [C(35, k) * (-4)^(35 - k) * (x^6)^(35 - k) * (-2)^k * (1/x)^k] (from k=0 to 35)

    = ∑ {C(35, k) * (-4)^(35 - k) * (-2)^k * x^[6(35 - k)] * x^(-k)] (from k=0 to 35)

    = ∑ {C(35, k) * (-4)^(35 - k) * (-2)^k * x^[6(35 - k) - k]} (from k=0 to 35).

    Now, the constant term's x exponent will be 0, so setting the exponent of x equal to 0 yields that the constant term occurs when:

    6(35 - k) - k = 0 ==> k = 30.

    Therefore, by evaluating C(35, k) * (-4)^(35 - k) * (-2)^k * x^[6(35 - k) - k] at k = 30, the constant term is:

    C(35, 30) * (-4)^(35 - 30) * (-2)^30 * x^0 = C(35, 30) * 2^40.

    I hope this helps!

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  • 7 years ago

    According to the Binomial Theorem the k-th term is C(35,k) (-4x^6)^k * (-2/x)^(35 - k)

    The power of x is (x^6)^k * (x^-1)^(35 - k) = x^(6k) * x^(-35 + k) = x^(6k - 35 + k).

    To be a constant term, this must be x^0. Set that expression equal to 0, solve for k, then plug k into the Binomial Expansion above.

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